Question: Divide the following complex numbers. $ \dfrac{20-21i}{5+2i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${5-2i}$ $ \dfrac{20-21i}{5+2i} = \dfrac{20-21i}{5+2i} \cdot \dfrac{{5-2i}}{{5-2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(20-21i) \cdot (5-2i)} {(5+2i) \cdot (5-2i)} = \dfrac{(20-21i) \cdot (5-2i)} {5^2 - (2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(20-21i) \cdot (5-2i)} {(5)^2 - (2i)^2} = $ $ \dfrac{(20-21i) \cdot (5-2i)} {25 + 4} = $ $ \dfrac{(20-21i) \cdot (5-2i)} {29} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({20-21i}) \cdot ({5-2i})} {29} = $ $ \dfrac{{20} \cdot {5} + {-21} \cdot {5 i} + {20} \cdot {-2 i} + {-21} \cdot {-2 i^2}} {29} $ Evaluate each product of two numbers. $ \dfrac{100 - 105i - 40i + 42 i^2} {29} $ Finally, simplify the fraction. $ \dfrac{100 - 105i - 40i - 42} {29} = \dfrac{58 - 145i} {29} = 2-5i $